Template type deduction and specialization

[Namespace via Digitalmars-d-learn]

On Wednesday, 20 May 2015 at 06:31:13 UTC, Mike Parker wrote:
> I don't understand why this behaves as it does. Given the 
> following two templates:
>
> ```
> void printVal(T)(T t) {
> 	writeln(t);
> }
> void printVal(T : T*)(T* t) {
> 	writeln(*t);
> }
> ```
>
> I find that I actually have to explicitly instantiate the 
> template with a pointer type to get the specialization.
>
> ```
> void main() {
> 	int x = 100;
> 	printVal(x);
> 	int* px = &x;
> 	printVal(px);        // prints the address
>         printVal!(int*)(px)  // prints 100
> }
> ```
>
> Intuitively, I would expect the specialization to be deduced 
> without explicit instantiation. Assuming this isn't a bug (I've 
> been unable to turn up anything in Bugzilla), could someone in 
> the know explain the rationale behind this?

What about:
----
import std.stdio;

void printVal(T)(T t) {
     static if (is(T : U*, U))
         printVal(*t);
     else
         writeln(t);
}

void main() {
     int x = 100;
     printVal(x);
     int* px = &x;
     printVal(px);
}
----